Can Gauss Curvature Be Zero on a Genus 2 Surface?

[Poirot1]

Use the gauss bonnet theorem to show that the gauss curvature of a closed orientable surface of genus 2 cannot be identically zero

euler characteristic is 2-2(2)=-2 so total gauss curvature is equal to -4pi. The integral of zero is zero and not -4pi so gauss curvature is not identically zero. Is this right?

 

[TheBigBadBen]

Use the gauss bonnet theorem to show that the gauss curvature of a closed orientable surface of genus 2 cannot be identically zero

euler characteristic is 2-2(2)=-2 so total gauss curvature is equal to -4pi. The integral of zero is zero and not -4pi so gauss curvature is not identically zero. Is this right?

That's right.

To explain this more clearly to those who end up at this thread:

The Gauss-Bonnet theorem states that for a compact, 2-dimensional manifold \(\displaystyle M\), with gaussian curvature \(\displaystyle K\), whose boundary \(\displaystyle ∂M\) has geodesic curvature \(\displaystyle k_g\), we have the equality:
\(\displaystyle \int_M K\;dA+\int_{\partial M}k_g\;ds=2\pi\chi(M)\)
where \(\displaystyle \chi(M)\) is the Euler Characteristic of the surface \(\displaystyle M\).

In the case of a closed, orientable surface of genus 2, we would state that the surface has no boundary, and that its Euler Characteristic is given by \(\displaystyle 2-2g=-2\). Thus, the Gauss-Bonnet theorem states that
\(\displaystyle \int_M K\;dA=-4\pi\)

Clearly, if \(\displaystyle K=0\) everywhere (that is, if K is "identically zero"), then the integral on the left would be the integral of zero over some surface, which would have to be zero. We conclude by contradiction that this cannot be the case.